In acute triangle $ABC$, points $D$, $E$, and $F$ are located on sides $\overline{BC}$, $\overline{AC}$, and $\overline{AB}$ so that $\overline{AD} \perp \overline{BC}$, $\overline{DE} \perp \overline{AC}$, and $\overline{DF} \perp \overline{AB}$. Let $R_1$ and $R_2$ be the radii of the circles around $\triangle ABC$ and $\triangle AEF$, respectively. Determine the number of degrees in the measure of $\angle A$ if the area of $\triangle ABC$ is equal to $R_1R_2$.
Let's start by drawing a diagram of the given situation. We have an acute triangle $ABC$ with points $D$, $E$, and $F$ located on sides $\overline{BC}$, $\overline{AC}$, and $\overline{AB}$ respectively, such that $\overline{AD}$ is perpendicular to $\overline{BC}$, $\overline{DE}$ is perpendicular to $\overline{AC}$, and $\overline{DF}$ is perpendicular to $\overline{AB}$.
To find the measure of angle $\angle A$, we need to identify a relationship between the area of triangle $ABC$ and the radii $R_1$ and $R_2$.
Let $O$ be the circumcenter of triangle $ABC$. It is a well-known fact that the circumcenter is the intersection point of the perpendicular bisectors of the sides of a triangle. Since $O$ is the circumcenter of triangle $ABC$, it lies on the perpendicular bisectors of segments $\overline{AC}$ and $\overline{AB}$.
Now, let's draw the circumcircle of triangle $AEF$ with center $O'$ and radius $R_2$. Since $O'$ lies on the perpendicular bisector of $\overline{AC}$, $O'$ must also be the midpoint of segment $\overline{AC}$. Similarly, $O'$ must be the midpoint of segment $\overline{AB}$.
Let's consider the right triangles $\triangle ADF$ and $\triangle ADE$. Since $\overline{DF} \perp \overline{AB}$ and $\overline{DE} \perp \overline{AC}$, these triangles are similar to $\triangle ABC$ by altitude similarity. This means that the ratios of corresponding side lengths are equal.
We have $\frac{DF}{AB} = \frac{DE}{AC} = \frac{DF+DE}{AB+AC}$. But notice that $DF+DE$ is the diameter of the circumcircle of triangle $AEF$, which is $2R_2$.
Hence, $\frac{DF+DE}{AB+AC} = \frac{2R_2}{AB+AC}$. Since triangle $ABC$ is acute, we can extend $\overline{AD}$ to $O$, the circumcenter of triangle $ABC$. Triangle $ABC$ is similar to triangle $AOD$ by SAS similarity.
Thus, we have $\frac{AO}{AB+AC} = \frac{OD}{AB} = \frac{DF+DE}{AB+AC} = \frac{2R_2}{AB+AC}$. Rearranging this equation gives us $AO = \frac{2R_2}{AB+AC} \cdot (AB+AC)$.
Now, let's consider the area of triangle $ABC$. We know that the area of a triangle can be expressed as $\frac{1}{2} \cdot AB \cdot AC \cdot \sin(\angle BAC)$. We can also express the area as $\frac{1}{2} \cdot AB \cdot OC \cdot \sin(\angle BAC)$, where $OC$ is the distance from the origin $O$ to side $\overline{BC}$.
Since triangle $ABC$ is acute, $O$ lies inside triangle $ABC$. This means that $OC$ is less than $\frac{1}{2} \cdot AB$. Let's express $OC$ as a fraction of $AB$ and substitute it into the area expression.
We have $OC = \frac{AO}{AB+AC} = \frac{2R_2}{AB+AC}$. Substituting this into the area expression gives us $\frac{1}{2} \cdot AB \cdot \frac{2R_2}{AB+AC} \cdot \sin(\angle BAC)$. Simplifying this expression gives us $R_2 \sin(\angle BAC)$.
Since the area of triangle $ABC$ is given as $R_1R_2$, we can equate these two expressions. This gives us $R_1R_2 = R_2 \sin(\angle BAC)$. We can cancel out $R_2$ from both sides, which gives us $R_1 = \sin(\angle BAC)$.
Now, we need to find the number of degrees in the measure of $\angle A$ given that $R_1 = \sin(\angle BAC)$. Since the range of the sine function is $[-1,1]$, we know that angle $\angle BAC$ must be an acute angle whose sine is $R_1$.
To find the number of degrees in the measure of $\angle A$, we can use the arcsine function. The arcsine function, denoted as $\arcsin(x)$, gives us the angle whose sine is $x$. Taking the arcsine of both sides of the equation $R_1 = \sin(\angle BAC)$ gives us $\angle A = \arcsin(R_1)$.
Hence, to find the number of degrees in the measure of $\angle A$, we need to calculate the arcsine of $R_1$.